 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2/32d/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Put

y = λ x

and

t = y + s

. Factoring out the endpoint value gives

γ (x, y) = y^{x - 1} e^{- y} \int_{0}^{\infty} exp ((x - 1) lo g (1 + \frac{s}{λ x}) - s) d s .

For fixed

s

(x - 1) lo g (1 + \frac{s}{λ x}) - s = - \frac{λ - 1}{λ} s - \frac{1}{x} (\frac{s}{λ} + \frac{s ^{2}}{2 λ ^{2}}) + O (x^{- 2}) .

Watson's endpoint lemma justifies termwise integration, so with

α = (λ - 1) / λ

\frac{γ ( x , y )}{y ^{x - 1} e ^{- y}} \sim \int_{0}^{\infty} e^{- α s} [1 - \frac{1}{x} (\frac{s}{λ} + \frac{s ^{2}}{2 λ ^{2}})] d s = \frac{1}{α} - \frac{1}{x} (\frac{1}{λ α ^{2}} + \frac{1}{λ ^{2} α ^{3}}) .

Consequently

f (λ) = \frac{λ}{λ - 1}, g (λ) = - \frac{λ ^{2}}{( λ - 1 ) ^{3}} .

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